KINETIC MOLECULAR MODEL AND INTERMOLECULAR FORCES

KINETIC MOLECULAR MODEL
AND INTERMOLECULAR FORCES
Kinetic Molecular Model of
Liquids and Solids

Recalling the concept:
Matter in the liquid state has indefinite
shape and definite volume.
Matter in the gas state has indefinite shape
and volume.
Matter in the solid state has definite shape
and volume.

SOLID
ARRANGEMENT OF PARTICLES
-closely and orderly packed
KINETIC ENERGY OF PARTICLES
-vibrate and rotate about a fixed
position
PARTICLE MOTION
-very low
ATTRACTIVE FORCES
-very strong

LIQUID
-less closely packed
-particles slide over each other
PARTICLE MOTION
-low
ATTRACTIVE FORCES
-strong

GAS
-very far apart
-particles move at great speed
PARTICLE MOTION
-high
ATTRACTIVE FORCES
-low

Polar attraction as a universal law…
 Molecules are held together by an
electrostatic attraction:
a. Intramolecular attraction
IONIC COVALENT
METALLIC

Polar attraction as a universal law…
 Molecules are held together by an
electrostatic attraction:
a. Intermolecular forces

van der Waals forces:
 The term for all known Intermolecular
forces.
 Named after a Dutch scientist:
Johannes van der Waals (1837 –
1932)

Types of van der Waals forces:
Ion – dipole
 Results when an ion and the partial charge
found at the end of the polar molecule attract
each other.
 Positive ions are attracted to the negative
end of a dipole and vice versa.
Example:
a. Salt (NaCl) Dissolved in Water (H2O)
b. Potassim (K+) Dissolved in Hydrochloric
Acid (HCl)

Ion – dipole
 Results when an ion and the partial charge
found at the end of the polar molecule attract
each other.
Example:
a. Salt (NaCl) Dissolved in Water (H2O)
b. Potassim (K+) Dissolved in Hydrochloric
Acid (HCl)

Dipole – dipole
 Exists between neutral polar molecules
 Polar molecules attract each other when the
positive end of one molecule is near the
negative end of another.
 Weaker force compared to ion-dipole
(depending on size)

Dipole – dipole
Example:
a. Dichloromethane
b. Hydrochloric Acid

London Dispersion Forces
 Force of attraction between nonpolar
molecules or atoms (Cl2 and CH4)
 Originated from Fritz London (1900-1954), a
German-American physicist

Instantaneous
dipole
Induced
dipoles

 Dipole can be induced more likely on
molecules having larger molecular masses.
(Polarizability)
-This also affects the melting and boiling
points of the molecules.

Hydrogen Bonds
 Plays an important role in life
processes
 It can easily be broken and reformed
 Occurs in water, DNA molecules and
protein

Hydrogen Bonds
 It is an attractive interaction between a
hydrogen atom bonded to an electronegative
Fluorine, Oxygen and Nitrogen atom and an
unshared electron pair of another nearby
electronegative atom.

Hydrogen Bonds
Example:
a. Water (H2O)
b. Ammonia
c. Ammonia and Water (NH3)
d. Hydrofluoric Acid (HF)

PROPERTIES OF LIQUIDS
What factors determine the physical
properties of liquids?

POPERTIES OF LIQUIDS
A. VISCOSITY
 What is the difference between fluid
and viscous liquids?
 VISCOSITY is the ability of a fluid to
resist flowing.
 Viscosity of a
liquid depends on
intermolecular
forces that is
present.

A. VISCOSITY
 Non-polar molecules have low
viscosities because of weak London
Force. Example: Benzene, pentane and
carbon tetrachloride.
 Polar molecules
such as glycerol
and aqueous
sugar solution
have high
viscosities.

A. VISCOSITY
What do you think is the effect of an
increasing temperature to the viscosity of
a liquid?
The
viscosity
decreases
as the

A. VISCOSITY
 VISCOMETER is a device used to
measure viscosity.

B. SURFACE TENSION

B. SURFACE TENSION
 The measure of the resistance of a
liquid to spread out.
 The higher the
temperature, the
less the strength
of the attractive
force that holds
the molecule
together

C. CAPILLARITY
 The rising of any liquid
 Results from
competition between
liquid’s intermolecular
force and the walls of
the tube.
 Capillarity is also
observed in plants’
transport system.

D. EVAPORATION, VAPOR
PRESSURE AND BOILING POINT
 Molecules of liquids, when obtained
enough kinetic energy liberates
 The escape of energetic molecules
in liquid reduces the average kinetic
energy of the remaining molecules

 What do you think is the relationship of
liquid evaporation to temperature and
pressure?
 The escape of energetic molecules
in liquid reduces the average kinetic
energy of the remaining molecules

Dynamic
Equilibrium
Vaporization Condensation

ANALYSIS
 Normal boiling point happens when a
liquid reaches an internal temperature
of 100OC under
1 atm (atmospheric pressure)
Which level with respect to sea
level foods cooks faster and
slower?

At higher altitude,
atmospheric pressure is
lesser.
Thus water boils faster at a
lower temperature because
less pressure is exerted on
water molecules.
Inefficient delivery of heat to
cook the food and it takes time
for the food to be cooked.

Lets check your
understanding.
Answer the following questions briefly in a ½ sheet of
pad paper; copy and answer.
1. Why do droplets of water come in spherical
shape on top of the leaves of the plants like
gabi?
2. Boiling points varies with location.
3. Your arm feels cool when alcohol
evaporates from your skin.
4. On a warm day, water droplets form on the
outside of the bottle of a carbonated
beverage.

PROPERTIES OF SOLIDS
How are the structures and properties of
solids related?

POPERTIES OF SOLIDS
 A SOLID is formed when
the temperature of a
liquid is low and the
pressure is sufficiently
high causing the
particles to come very
close to one another.
 They are rigid
 Their particles hardly
diffuse

POPERTIES OF SOLIDS
NATURE OF SOLIDS
Crystalline
Solids
Amorphous
Solids

POPERTIES OF SOLIDS
A. CRYSTALLINE
 Atoms, ions, or molecules are arranged
in well defined arrangement
 Having flat surface and sharp edges
 Example: gems, salts, sugar and ice.

POPERTIES OF SOLIDS
TYPES OF CRYSTALLINE SOLIDS
Ionic
Molecular
Covalent Network
Metallic

POPERTIES OF SOLIDS
1. Ionic Crystalline Solids
 Composed of (+) and (-) ions
 Held by electrostatic attractions
 They are hard, brittle and poor
electrical and thermal conduction
 Example: NaCl

POPERTIES OF SOLIDS
2. Molecular Crystalline Solids
 Composed of atoms and molecules
 Held together by: H-Bond, dipole-
dipole, and London dispersion forces
 Soft, low to moderate melting point and
poor thermal and electrical
conductivity
 Examples: CH4, C12H22O11, CO2, H2O
and Br2

POPERTIES OF SOLIDS
3. Covalent Network Crystalline
Solids
 Atoms connected in a network of
covalent molecules
 Held together by covalent bonds
 Very hard, very high melting point and
often poor thermal and electrical
conductivity.
 Examples: Plastics, Allotropes of
carbon, silicon carbide

POPERTIES OF SOLIDS
4. Metallic Crystalline Solids
 Composed of atoms and molecules
 Held together by metallic bonds
 Soft to hard, low to high melting point,
malleable, ductile and good thermal
and electrical conduction
 All metallic elements: Cu, Na, Zn, Fe
and Al

POPERTIES OF SOLIDS
How are molecules being
arranged in microscopic level?
Unit Cell
Crystal
Lattice
The smallest portion of the
crystal which shows the
complete pattern of its
particles
The repetition of unit cells
in all directions

POPERTIES OF SOLIDS
Structural
Representations
of Molecules:
1. NaCl
2. Ice
3. Diamond
4. Metallic Bond
in (Fe)

POPERTIES OF SOLIDS
B. AMORPHOUS SOLIDS
 From the Greek word for “without
form”
 Solid particles which do not have
orderly structures.
 They have poorly defined shapes
 are rigid, but they lack repeated
periodicity or long-range order in their
structure.
 examples include thin film lubricants,
metallic glasses, polymers, and gels

POPERTIES OF SOLIDS
STRUCTURE OF AMORPHOUS SOLID:

POPERTIES OF SOLIDS
CRYSTALLINE VERSUS AMORPHOUS SOLIDS

POPERTIES OF SOLIDS
 It can be noted that as temperature of
crystalline solid is increased, the
particles vibrate back and forth about
its lattice point.
 The crystal becomes less ordered.
 The heat added increases the kinetic
motion of the particles.
 Until the crystalline structure is
completely destroyed by the vibrations
of the particles, melting is achieved.

ANALYSIS
 What will happen if heating stops and
no heat is allowed to escape?
Both solid and liquid phases are
present in equilibrium.

PHASE CHANGE
Heating curve for the conversion of ice to
gaseous water.

PHASE CHANGE
ANALYSIS
 How does intermolecular force relates
to the rate at which melting point of a
substance is achieved?
Forces of attraction are weak in
substances with lower melting
point and vice versa.

PHASE CHANGE
VOCABULARIES
 HEAT FUSION- refers to the amount of
energy required to overcome the
intermolecular forces to convert a solid
into a liquid
 HEAT VAPORIZATION- the amount of
energy required to convert a liquid into
a gas.

PHASE CHANGE
SYNTHESIS
 Ionic Compounds have very high
melting point because of a very strong
intermolecular force.
 Example:
 NaCl
 MgCl2
 BeF
 CaF2

PHASE CHANGE
SYNTHESIS
 Covalent compounds have low to
moderate melting point because of
weak intermolecular force.
 Example:
 Water
 Glycerin
 Hormones
 Other Fats

QUANTITATIVE ASPECTS IN PHASE
CHANGES
 Different substances absorbs heat in
varying amounts.
 SPECEFIC HEAT is defined as the amount
of heat needed to raise the temperature of
one gram of substance by one degree
Celsius.
Q = mc T
Q= heat
m= mass
c= specific heat capacity
T= change in temperature

CHANGES
NOTE:
 When materials with small specific heat
value absorbs energy, its temperature rises
rapidly.
 In contrast, materials with high specific heat
values absorb a large amount of heat
without much increase in temperature.
 Water has a specific heat capacity of 4.16
Joules.

CHANGES
SAMPLE PROBLEM:
 Hot water at 100oC can burn and damage
the skin, but the effect of steam on the skin
can be even more severe. Calculate the
amount of heat absorbed by the skin from a
150-g steam burned at 100oC.

ENERGY CHANGES IN
CHEMICAL REACTIONS
Thermochemistry

THERMOCHEMISTRY
INTRODUCTION
Energy is the foundation of the
universe.

THERMOCHEMISTRY
 Energy transfer may be in the form of
heat or work.
 HEAT (Q) – is the transfer or energy
between a system and surroundings
due to temperature difference.
 Heat may be absorbed or released by a
system depending on which has a
higher temperature between the
system and the surroundings.

THERMOCHEMISTRY
 During chemical reaction, there is an
energy change between molecules.
 TWO TYPES OF REACTIONS:
Endothermic
Exothermic

THERMOCHEMISTRY
THERMOCHEMISTRY
 The study of energy changes that
occur during chemical reactions and
changes of state.

THERMOCHEMISTRY
ANALYSIS
 How does energy undergo change
within a system or within a chemical
reaction?
Heat flows in and out of the
system during chemical reactions.

THERMOCHEMISTRY
THE LAW OF CONSERVATION OF
ENERGY:
 In any chemical or physical process,
energy is neither created nor
destroyed.
 In any chemical or physical process,
energy in the universe remains
unchanged.
Energyuniv = constant

THERMOCHEMISTRY
Recitation: Explain how energy is
conserved in the following situations:
 Burning of gasoline
 Hydroelectric powerplant
 Cellphone telecommunications
 Condensation of water vapor
 Induction cooking

THERMOCHEMISTRY
THREE TYPES OF SYSTEMS:
Open System
Closed System
Isolated System

THERMOCHEMISTRY
OPEN SYSTEM
 Matter and energy occurs between
system and surrounding
 System interacts with the surrounding

THERMOCHEMISTRY
CLOSED SYSTEM
 Only energy can transfer between
system and surroundings
ISOLATED SYSTEM
 Matter and energy cannot transfer
between the system and its
surroundings.
 Example: contents of adiabatic bomb
calorimeter.

THERMOCHEMISTRY
The magnitude of heat can be computed
using the following equation
Q = mc T
Q= heat
m= mass
c= specific heat capacity
T= change in temperature

THERMOCHEMISTRY
WORK (w)
 Force applied over a given distance
 Energy transfer between a system and
the surrounding due to a force acting
through a distance

THERMOCHEMISTRY
TABLE 3.2: Assigned Convention for Work, w
System does work
on the
surroundings
-w Expansion
Surroundings does
work on the system +w Compression

THERMOCHEMISTRY
SAMPLE PROBLEM:
 How much work is needed in a system
to expand from 25 to 50 liters against a
pressure of 5 atm? Is work done by the
system or on the system?

THERMOCHEMISTRY
SAMPLE PROBLEM:
 How much work is needed in a
system to compress a carbon
dioxide gas inside a fire
extinguisher from the volume of
500 liter to 275 liter at 3.5 atm? Is
work done by the system or on the
sytem?

THERMOCHEMISTRY
ENTHALPY
 In a chemical reaction, there is an
energy change from the beginning up
to the end of the reaction.
 Change in energy: ENDOTHERMIC OR
EXOTHERMIC REACTION
 Represented by H

THERMOCHEMISTRY
ENTHALPY
 Energy change in the reaction or the
sum of all the energy stored in the
bonds of the product minus the energy
stored in the bond of the reactant
 If there is more energy in the product
than the reactant, the value of H is
positive = ENDOTHERMIC REACTION

THERMOCHEMISTRY
THERMOCHEMICAL EQUATIONS
 If there is less energy in the product than
the reactant, the value of H is
negative = EXOTHERMIC REACTION

THERMOCHEMISTRY
WRITING THERMOCHEMICAL
EQUATIONS:
I. When heat is lost, the H value is
negative. (Exothermic reaction)
CH4(g) + 2O2 ------ CO2(g) + 2H2O(l)
H = -890 kJ

THERMOCHEMISTRY
EQUATIONS:
I. When heat is gained, the H value is
positive. (Endothermic reaction)
CO2(g) + 2H2O(l) ------ CH4(g) + 2O2
H = 890 kJ

THERMOCHEMISTRY
EQUATIONS:
II. HEAT IS A STATE FUNCTION, thus
energy changes or H value for the same
equation may be different if it occurs in
different physical state.

THERMOCHEMISTRY
EQUATIONS:
III. If a reaction is reversed, then the
enthalpy H value will also be reversed.
Hence +a becomes –a and vice versa

THERMOCHEMISTRY
EQUATIONS:
III. If we change the stoichiometric
coefficients in the reaction, we also
change the enthalpy H value
proportionally!

THERMOCHEMISTRY
EQUATIONS:
a. If coefficient is doubled enthalpy must be
doubled also
b. If we triple the coefficient, enthalpy must
also be tripled
c. Same with when we half the coefficient.
d. If we double the equation and reverse, we
must also double the enthalpy and
reverse the sign

THERMOCHEMISTRY
Sample Problem:
a. Manipulate the thermochemical equation
below as endothermic reaction:
CH4(g) + 2O2 ------ CO2(g) + 2H2O(l)
H = -890 kJ

THERMOCHEMISTRY
HESS LAW
 States that the enthalpy change of an overall
reaction is the sum of the enthalpy changes
of its individual steps.
EXAMPLE: we can burn carbon directly to
carbon dioxide
C(s) + O2(g) ------------ CO2(g) H= -393.509 kJ
OR

THERMOCHEMISTRY
HESS LAW
 Carbon to carbon monoxide, then carbon
monoxide to carbon dioxide
C(s) + ½ O2(g) ----------- CO(G) H= -110.524 kJ
CO(g) + ½ O2(g) --------- CO2(g) H= -282.958 kJ
____________________________________________
C(s) + O2(g) ------------- CO2(g) H= -393.509 kJ

THERMOCHEMISTRY
STEPS IN GETTING THE HEAT
SUMMATION:
1. Identify the net equation whose ∆H is
unknown. Make sure that the reaction is
balanced.
2. Manipulate the equations where ∆H is
known so that the correct moles of the
reactants and the products are on correct
sides.

THERMOCHEMISTRY
STEPS IN GETTING THE HEAT
SUMMATION:
1. Add these individual reactions to get the net
reaction. The value of the unknown ∆H is the
sum of the individual manipulated ∆H.

REACTION RATES
Chemical Kinetics

Chemical Kinetics
Reaction Rate
 How fast the reaction takes place
 Some reactions proceed at very fast rate
while others proceed very slowly
 Fractions of the reactants are changed into
product until all the substances are
converted fully.
(R)Rate= M(molar mass)/s(second)

Examples of Reaction
 Burning of Rocket Fuel
 Spoiling of food outside the refrigerator
 Rusting of Iron
Chemical Kinetics

Factors Affecting Rates of Reactions
A. Nature of Reactants:
 Reaction depends on the particular
reactants and the number of bonds that
have to be broken.
 Reactions are rapid between oppositely-
charged particles
 Reactions involving covalent substances
are slow at room temperature.
 Gasses proceed quick reactions than solid
and liquid.
 The reactions between heterogeneous
reactions are slower than homogeneous
reactions

B. Concentration:
 An increase in concentration of the reactant
indirectly means an increase in collision
theory, thus increasing the reaction rate.

C. Surface Area:
 The smaller the size of particles, the larger
the surface area exposed.
A larger surface area increases the frequency
of collisions

C. Effects of Catalyst:
 Provides an alternative pathway of lower
activation energy.
 Representation of a chemical equation with
the presence of catalyst:
2H2O2 → O2 + 2H2O
MnO4

C. Effects of Temperature
 Food spoilage at room temperature on warm
summer days
 Most plants grow faster in warm than in cold
weather.
 Animals living at moderate pressure under
the deep sea are fatty than those fishes
living in the shallow portion of the
sea/freshwater.

Identify the factors that influence
reaction rates and explain:
1. A brush (grass fire) spreads more rapidly
on a sunny day than on an overcast day.
2. Sodium reacts more rapidly with water than
iron does.
3. Powdered zinc reacts more rapidly with
sulfuric acid than a large piece of zinc of
equal weight does.
4. It is more dangerous to drop a lighted
match into a gasoline tank that has just
been emptied than into one which is
completely full.
5. Cake batter will cook only when heated.

QUIZ
Identify the factors that influence reaction
rates and explain:

PROGRESS OF
CHEMICAL REACTION
Chemical Reaction

THERMOCHEMISTRY
RATE LAW
 The concentration of reactants influences
the rate of chemical reactions.
 The effect of concentration of reactants on
the rate of reaction can be seen
quantitatively using the rate law for the
reaction.
 An expression that gives a mathematical
relationship of the rate of a reaction and the
concentration of reactants.
Rate= k[A]m[B]n

THERMOCHEMISTRY
Consider the equation between oxygen
and nitric oxide in the formation of acid
rain:
O2(g) + 2NO(g) → 2NO2(g)
Rate= k[O2]m[NO]n

THERMOCHEMISTRY
Rate= k[O2]m[NO]n
 Brackets represent the concentration of the
reactants given in moles per liter
 k= the fixed value for rate constant
 m and n represent the order of the reaction
 In getting the experimental value for m and
n, the concentration of one of the reactants
is changed while the other is kept constant.

THERMOCHEMISTRY
 Doubling the
initial
concentration
of the
reactants, the
rate of the
reaction is also
doubled
The reaction is FIRST ORDER
WITH RESPECT TO N2O5

THERMOCHEMISTRY
Example Problem:
R= k[O2][NO]2

THERMOCHEMISTRY
R= k[O2][NO]2
0.04M/s=k(0.250M)(0.100M)
k= 1.6 M/s

Rate of Chemical Reaction
The following data were obtained for the
reaction:
A2 + B2 +3C → D
Experiment
No.
Reactant Concentrations Rate of
Reaction
(M/s)
[A] [B] [C]
1 0.10 0.20 0.20 0.0090
2 0.20 0.20 0.30 0.0360
3 0.20 0.60 0.30 0.0720
4 0.20 0.20 0.60 0.2880

BALANCING REDOX REACTIONS
Review:
 Electrons makes it possible for one atom
bind with another atom.
 In acid-base reactions, transfer of proton
(H+) is involved.

What is a REDOX reaction?
 RED(Reduction): Substance gain an
electron
 Antoine Lavosier may leave or goes in into a
substance, thus changing its mass.
 OX(Oxidation): Tendency of a substance to
loose an electron.

REDOX IN NATURE:
 Cellular Respiration
 Photosynthesis
 Battery charging
 Burning process

In electron transfer,
 We track electrons like a banking
transactions.
Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)
Fe(s) + Cu2+
(aq) + SO4
2-
(aq) → Fe2+
(aq) + SO4
2-
(aq) + Cu(s)
Fe(s) + Cu2+
(aq) → Fe2+
(aq) + Cu(s)
Lost of e- (oxidized)
Gain of e- (reduced)
Reducing Agent
Oxidizing Agent

Oxidation number in tracking electrons:
 Hypothetical value for each atom in a
molecule (not actual)
 ELEMENT: zero (0) oxidation# (He, O2, Fe)
 MONOATOMIC ION: the same with its ionic
charge (Oxygen -2, Fe +2)
 NEUTRAL MOLECULE: ox. Numbers add up
to get zero [CO2 (+4 -4)]

Assigning Oxidation Number:
4Fe(s) + 3O2 → 2Fe2O3
0 0 +3 -2
Fe: 0 → +3: Loss of –e [oxidized]
O: 0 → -2: Gain of –e [reduced] Oxidizing Agent
Reducing Agent

Practice
PbO2
MnO4
S
NH3

Balancing a Redox Reaction
0 0 +3 -2
Fe: 0 → +3: Loss of –e [oxidized]
O: 0 → -2: Gain of –e [reduced] Oxidizing Agent
Reducing Agent

Consider the aqueous solution iron (II) ion (Fe3+) with
dichromate ion (Cr2O7
-2):
4Fe2+ + Cr2O7
-2 → 4Fe3+ Cr3+
STEPS:
Separate the unbalanced reaction into half-reactions. A half
reaction is an oxidation/reduction that occurs as part of
overall redox reaction.
Oxidation: Fe2+ → Fe3+
Reduction: Cr2O7
-2 → Cr3+

4Fe2+ + Cr2O7
-2 → 4Fe3+ Cr3+
STEPS:
Balance each of the half-reactions with regard to atoms other
than O and H. In this case, no change is required for the
oxidation half-reaction. We adjust the coefficient of the
chromium (III) ion to balance the reduction half reaction.
Reduction: Cr2O7
-2 → 2Cr3+

4Fe2+ + Cr2O7
-2 → 4Fe3+ Cr3+
STEPS:
Balance both half-reactions for H by adding H+ . Once again,
the oxidation in this case requires no change, but we must add
14 hydrogen ions to the product side of the reaction.
Reduction: Cr2O7
-2 → 2Cr3+ + 7H2O

4Fe2+ + Cr2O7
-2 → 4Fe3+ Cr3+
STEPS:
Balance both half-reactions for H by adding H+ Once again,
the oxidation in this case requires no change but we must add
14 hydrogen ions to the reactant side of the reaction
Reduction: 14H+ + Cr2O7
-2 → 2Cr3+ + 7H2O

4Fe2+ + Cr2O7
-2 → 4Fe3+ Cr3+
STEPS:
Balance both half-reactions for charge by adding electrons. To
do this, we determine the total charge on each side and add
electrons to make total charges equal.
Oxidation: Fe2+ → Fe3+ + e-
+2 +2

4Fe2+ + Cr2O7
-2 → 4Fe3+ Cr3+
STEPS:
In case of reduction, there is total charge of
[(14)(+1) + (2)(-)] = +12 on the reactant side and a total charge
of [(2)(+3)] = +6 on the product side. Adding six electrons to
the reactant side makes the charges equal
Reduction: + 14H+ + Cr2O7
-2 → 2Cr3+ + 7H2O
+6 +6
6e-

4Fe2+ + Cr2O7
-2 → 4Fe3+ Cr3+
STEPS:
If the number of electrons in the balanced oxidation half-reaction
is not the same as the number of electrons in the balanced
reduction half-reaction, multiply one or both of the half-reactions
by the number(s) req. to make it balanced.
Oxidation: 6(Fe2+ → Fe3+ e-)
6Fe2+ → 6Fe3+ 6e-
Reduction: 6e- +14H+ + Cr2O7
-2 → 2Cr3+ + 7H2O

4Fe2+ + Cr2O7
-2 → 4Fe3+ Cr3+
STEPS:
Finally, add the balanced half-reactions back together and cancel
the electrons, in addition to any other identical terms that appear
on both sides.
6Fe2+ → 6Fe3+ 6e-
6e- +14H+ + Cr2O7
-2 → 2Cr3+ + 7H2O
6Fe2++14H+ + Cr2O7
-2 → 6Fe3+ + 2Cr3+ + 7H2O

KINETIC MOLECULAR MODEL AND INTERMOLECULAR FORCES

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